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never [62]
4 years ago
9

Please answer this question and I will make your answer a brainlist!!! This is super urgent!!! I will highly appreciate it.

Mathematics
1 answer:
dusya [7]4 years ago
8 0

Answer:

0?

Step-by-step explanation:

idk if its correct..but thats my ans..

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Use the graph of the quadratic, g(x) to answer each question
forsale [732]

Answer:

-5

Step-by-step explanation:

from graph

6 0
3 years ago
C=wtc/1000 solve for w
Nutka1998 [239]
For this case we have the following equation:
 C =  \frac{wtc}{1000}
 To clear w, we must follow the following steps:
 1) The value of t multiplied by c pass to divide the other side of the equation:
 \frac{w}{1000} =  \frac{C}{tc}
 2) the value of 1000 is passed to multiply to the other side of the equation:
 w = \frac{1000C}{tc}
 Answer:
 
The cleared equation for w is:
 
w = \frac{1000C}{tc}
3 0
3 years ago
Read 2 more answers
A model giraffe has a scale of 1 in : 3 ft. If the model giraffe is 4 in tall, then how tall is the real giraffe
VladimirAG [237]
The answer is 12ft. Because if 1in is 3ft then 4in will be 4 multiplied by 3
4 0
4 years ago
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X=5y: 2x+5y=60<br><br> A) 10,4<br> B) 15,4<br> C) 20,4<br> D) 25,4
elena-14-01-66 [18.8K]

Answer:

C) 20, 4

Step-by-step explanation:

X = 5y

2x + 5y = 60

substitute in 5y for x: 2(5y) + 5y = 60

10y + 5y = 60

15y = 60

y = 4

x = 5y

x = 5(4) = 20

3 0
3 years ago
Read 2 more answers
Lifetime of $1 Bills The average lifetime of circulated $1 bills is 18 months. A researcher believes that the average lifetime i
OLEGan [10]
<h2>Answer with explanation:</h2>

Let \mu be the population mean lifetime of circulated $1 bills.

By considering the given information , we have :-

H_0:\mu=18\\\\H_a:\mu\neq18

Since the alternative hypotheses is two tailed so the test is a two tailed test.

We assume that the lifetime of circulated $1 bills is normally distributed.

Given : Sample size :  n=50 , which is greater than 30 .

It means the sample is large so we use z-test.

Sample mean : \overline{x}=18.8

Standard deviation : \sigma=2.8

Test statistic for population mean :-

z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}

\Rightarrow\ z=\dfrac{18.8-18}{\dfrac{2.8}{\sqrt{50}}}\approx2.02

The p-value= 2P(z>2.02)=0.0433834

Since the p-value (0.0433834) is greater than the significance level (0.02) , so we do not reject the null hypothesis.

Hence, we conclude that we do not have enough evidence to support the alternative hypothesis that the average lifetime of a circulated $1 bill differs from 18 months.

6 0
3 years ago
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